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EE 5340 Semiconductor Device Theory Lecture 05 – Spring 2011

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Presentation on theme: "EE 5340 Semiconductor Device Theory Lecture 05 – Spring 2011"— Presentation transcript:

1 EE 5340 Semiconductor Device Theory Lecture 05 – Spring 2011
Professor Ronald L. Carter

2 Review the Following R. L. Carter’s web page:
EE 5340 web page and syllabus. (Refresh all EE 5340 pages when downloading to assure the latest version.) All links at: University and College Ethics Policies Makeup lecture at noon Friday (1/28) in 108 Nedderman Hall. This will be available on the web. ©rlc L05-08Feb2011

3 First Assignment Send e-mail to ronc@uta.edu
On the subject line, put “5340 ” In the body of message include address: ______________________ Your Name*: _______________________ Last four digits of your Student ID: _____ * Your name as it appears in the UTA Record - no more, no less ©rlc L05-08Feb2011

4 Second Assignment Submit a signed copy of the document posted at
©rlc L05-08Feb2011

5 Schedule Changes Due to University Weather Closings
Make-up class will be held Friday, February 11 at 12 noon in 108 Nedderman Hall. Additional changes will be announced as necessary. Syllabus and lecture dates postings will be updated in the next 24 hours. Project Assignment will be posted in the next 36 hours. ©rlc L05-08Feb2011

6 Intrinsic carrier conc. (MB limit)
ni2 = no po = Nc Nv e-Eg/kT Nc = 2{2pm*nkT/h2}3/2 Nv = 2{2pm*pkT/h2}3/2 Eg = 1.17 eV - aT2/(T+b) a = 4.73E-4 eV/K b = 636K ©rlc L05-08Feb2011

7 Classes of semiconductors
Intrinsic: no = po = ni, since Na&Nd << ni, ni2 = NcNve-Eg/kT, ~1E-13 dopant level ! n-type: no > po, since Nd > Na p-type: no < po, since Nd < Na Compensated: no=po=ni, w/ Na- = Nd+ > 0 Note: n-type and p-type are usually partially compensated since there are usually some opposite- type dopants ©rlc L05-08Feb2011

8 Equilibrium concentrations
Charge neutrality requires q(po + Nd+) + (-q)(no + Na-) = 0 Assuming complete ionization, so Nd+ = Nd and Na- = Na Gives two equations to be solved simultaneously 1. Mass action, no po = ni2, and 2. Neutrality po + Nd = no + Na ©rlc L05-08Feb2011

9 Equilibrium conc (cont.)
For Nd > Na (taking the + root) no = (Nd-Na)/2 + {[(Nd-Na)/2]2+ni2}1/2 For Nd >> Na and Nd >> ni, can use the binomial expansion, giving no = Nd/2 + Nd/2[1 + 2ni2/Nd2 + … ] So no = Nd, and po = ni2/Nd in the limit of Nd >> Na and Nd >> ni ©rlc L05-08Feb2011

10 n-type equilibrium concentrations
N ≡ Nd - Na , n type  N > 0 For all N, no = N/2 + {[N/2]2+ni2}1/2 In most cases, N >> ni, so no = N, and po = ni2/no = ni2/N, (Law of Mass Action is al- ways true in equilibrium) ©rlc L05-08Feb2011

11 Position of the Fermi Level
Efi is the Fermi level when no = po Ef shown is a Fermi level for no > po Ef < Efi when no < po Efi < (Ec + Ev)/2, which is the mid-band ©rlc L05-08Feb2011

12 p-type equilibrium concentrations
N ≡ Nd - Na , p type  N < 0 For all N, po = |N|/2 + {[|N|/2]2+ni2}1/2 In most cases, |N| >> ni, so po = |N|, and no = ni2/po = ni2/|N|, (Law of Mass Action is al- ways true in equilibrium) ©rlc L05-08Feb2011

13 Position of the Fermi Level
Efi is the Fermi level when no = po Ef shown is a Fermi level for no > po Ef < Efi when no < po Efi < (Ec + Ev)/2, which is the mid-band ©rlc L05-08Feb2011

14 EF relative to Ec and Ev Inverting no = Nc exp[-(Ec-EF)/kT] gives Ec - EF = kT ln(Nc/no) For n-type material: Ec - EF =kTln(Nc/Nd)=kTln[(Ncpo)/ni2] Inverting po = Nv exp[-(EF-Ev)/kT] gives EF - Ev = kT ln(Nv/po) For p-type material: EF - Ev = kT ln(Nv/Na) ©rlc L05-08Feb2011

15 EF relative to Efi Letting ni = no gives  Ef = Efi ni = Nc exp[-(Ec-Efi)/kT], so Ec - Efi = kT ln(Nc/ni). Thus EF - Efi = kT ln(no/ni) and for n-type EF - Efi = kT ln(Nd/ni) Likewise Efi - EF = kT ln(po/ni) and for p-type Efi - EF = kT ln(Na/ni) ©rlc L05-08Feb2011

16 Locating Efi in the bandgap
Since Ec - Efi = kT ln(Nc/ni), and Efi - Ev = kT ln(Nv/ni) The 1st equation minus the 2nd gives Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv) Since Nc = 2.8E19cm-3 > 1.04E19cm-3 = Nv, the intrinsic Fermi level lies below the middle of the band gap ©rlc L05-08Feb2011

17 Example calculations For Nd = 3.2E16/cm3, ni = 1.4E10/cm3
no = Nd = 3.2E16/cm3 po = ni2/Nd , (po is always ni2/no) = (1.4E10/cm3)2/3.2E16/cm3 = 6.125E3/cm3 (comp to ~1E23 Si) For po = Na = 4E17/cm3, no = ni2/Na = (1.4E10/cm3)2/4E17/cm3 = 490/cm3 ©rlc L05-08Feb2011

18 Sample calculations Efi = (Ec + Ev)/2 - (kT/2) ln(Nc/Nv), so at 300K, kT = meV and Nc/Nv = 2.8/1.04, Efi is 12.8 meV or 1.1% below mid-band For Nd = 3E17cm-3, given that Ec - EF = kT ln(Nc/Nd), we have Ec - EF = meV ln(280/3), Ec - EF = eV =117meV ~3x(Ec - ED) what Nd gives Ec-EF =Ec/3 ©rlc L05-08Feb2011

19 Equilibrium electron conc. and energies
©rlc L05-08Feb2011

20 Equilibrium hole conc. and energies
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21 vx = axt = (qEx/m*)t, and the displ
Carrier Mobility In an electric field, Ex, the velocity (since ax = Fx/m* = qEx/m*) is vx = axt = (qEx/m*)t, and the displ x = (qEx/m*)t2/2 If every tcoll, a collision occurs which “resets” the velocity to <vx(tcoll)> = 0, then <vx> = qExtcoll/m* = mEx ©rlc L05-08Feb2011

22 Carrier mobility (cont.)
The response function m is the mobility. The mean time between collisions, tcoll, may has several important causal events: Thermal vibrations, donor- or acceptor-like traps and lattice imperfections to name a few. Hence mthermal = qtthermal/m*, etc. ©rlc L05-08Feb2011

23 Carrier mobility (cont.)
If the rate of a single contribution to the scattering is 1/ti, then the total scattering rate, 1/tcoll is ©rlc L05-08Feb2011

24 Figure (p. 31 M&K) Electron and hole mobilities in silicon at 300 K as functions of the total dopant concentration. The values plotted are the results of curve fitting measurements from several sources. The mobility curves can be generated using Equation with the following values of the parameters [3] (see table on next slide). ©rlc L05-08Feb2011

25 Parameter Arsenic Phosphorus Boron
μmin 52.2 68.5 44.9 μmax 1417 1414 470.5 Nref 9.68 X 1016 9.20 X 1016 2.23 X 1017 α 0.680 0.711 0.719 Figure (cont. M&K) ©rlc L05-08Feb2011

26 Drift Current The drift current density (amp/cm2) is given by the point form of Ohm Law J = (nqmn+pqmp)(Exi+ Eyj+ Ezk), so J = (sn + sp)E = sE, where s = nqmn+pqmp defines the conductivity The net current is ©rlc L05-08Feb2011

27 Drift current resistance
Given: a semiconductor resistor with length, l, and cross-section, A. What is the resistance? As stated previously, the conductivity, s = nqmn + pqmp So the resistivity, r = 1/s = 1/(nqmn + pqmp) ©rlc L05-08Feb2011

28 Drift current resistance (cont.)
Consequently, since R = rl/A R = (nqmn + pqmp)-1(l/A) For n >> p, (an n-type extrinsic s/c) R = l/(nqmnA) For p >> n, (a p-type extrinsic s/c) R = l/(pqmpA) ©rlc L05-08Feb2011

29 References M&K and 1Device Electronics for Integrated Circuits, 2 ed., by Muller and Kamins, Wiley, New York, 1986. See Semiconductor Device Fundamen-tals, by Pierret, Addison-Wesley, 1996, for another treatment of the m model. 2Physics of Semiconductor Devices, by S. M. Sze, Wiley, New York, 1981. ©rlc L05-08Feb2011

30 References *Fundamentals of Semiconductor Theory and Device Physics, by Shyh Wang, Prentice Hall, 1989. **Semiconductor Physics & Devices, by Donald A. Neamen, 2nd ed., Irwin, Chicago. M&K = Device Electronics for Integrated Circuits, 3rd ed., by Richard S. Muller, Theodore I. Kamins, and Mansun Chan, John Wiley and Sons, New York, 2003. ©rlc L05-08Feb2011


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